| Original Message (ID '137897') By Bill Simpson: |
| Mathematica has never shown the steps to calculate a result, often even the algorithm used is undocumented.
Sometimes you can write a simple calculation yourself and learn something in the process.
From http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html P(x) a Lagrange polynomial is a sum of product/product.
If I haven't made any mistakes then this is one way of calculating that
In[1]:= p[points_]:=Sum[Product[If[i==j,1,x-points[[i,1]]],{i,1,Length[points]}]/ Product[If[i==j,1,points[[i,1]]-points[[j,1]]],i,1,Length[points]}] *points[[j,2]], {j,1,Length[points]}];
poly=p[{{1,1},{2,4},{3,2}}]
Out[2]=
1/2(-3+x)(-2+x)-4(-3+x)(-1+x)+(-2+x)(-1+x)
In[3]:= {x,poly}/.{{x->1},{x->2},{x->3}}
Out[3]= {{1,1},{2,4},{3,2}}
And so the same input x values create the same output y values.
You can also Expand that polynomial to get it into a more usual form if you like.
Please check this carefully on different sets of data until you are fairly sure I didn't make any misatkes. You can also compare that definition of p with the example on the web page and try to understand how the two are related. |
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